triangleABC BC=1/2(AB+AC)

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Source: Bosnia and Herzegovina 2011

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118 posts

RaleD

#1 h May 16, 2011, 10:19 AM • 2 Y

Y by Adventure10, Mango247

In triangle $ABC$ it holds $|BC|= \frac{1}{2}(|AB|+|AC|)$ . Let $M$ and $N$ be midpoints of $AB$ and $AC$ , and let $I$ be the incenter of $ABC$ . Prove that $A, M, I, N$ are concyclic.

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#2 h May 16, 2011, 1:01 PM • 3 Y

Y by Durjoy1729, Adventure10, Mango247

Let $S$ be circumcenter of triangle $ABC$ and let angle bisector at $A$ cut circumcircle of triangle $ABC$ at $A'$ .
Since $A,M,N$ and $S$ are concyclic with $\angle MSA = 90^{\circ}$ it is enought to prove that $I$ bisect $AA'$ .
But this could be easly proved by Ptolomy:

$AA'\cdot BC = AB\cdot A'C+AC\cdot A'B$

Since $A'I = A'B =A'C$ this follows.

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#3 h May 18, 2011, 2:21 AM • 1 Y

Y by Adventure10

From $BC= \frac{AB+AC}{2}$ we have that $BC = BM+CN$ . Let $E$ be the point in $BC$ such that $BM= CE$ and $CN = CE$ .

Since $BI$ and $CI$ are angle bisectors of angles $\angle ABC$ and $\angle ACB$ , respectively, and triangles $BME$ and $CNE$ are isosceles we have that $BI$ and $CI$ are the perpendicular bisectors of $ME$ and $NE$ , respectively. Hence $I$ is the circumcenter of triangle $MEN$ .

Therefore, since $I$ lies on the perpendicular bisector of $MN$ and on the angle bisector of $\angle MAN$ we have that $AMIN$ is cyclic.

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jgnr

#4 h May 18, 2011, 5:43 PM • 2 Y

Y by Adventure10, Mango247

Let AI meets BC at D and the circumcircle of ABC at K. We have AB/BD = c/(ac/(b+c)) = 2. Since ABD is similar to AKC, then AK=2KC. This also implies AK=2KI. The homothety centered at A of ratio 1/2 brings K,B,C to I,M,N, and also the circumcircle of ABC to the circle passing through A,M,I,N, so they are concyclic, as desired.

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#5 h May 18, 2011, 7:26 PM • 2 Y

Y by Adventure10, Mango247

Btw, I had this problem on my geometry test..

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#6 h May 28, 2011, 11:29 AM • 2 Y

Y by Adventure10, Mango247

master_Hjom wrote:

Btw, I had this problem on my geometry test..

We know that $AIXZ$ is cyclic, where $X,Z$ are the points of perpendiculars from $I$ to $AB$ and $AC$ respectively. So we need to prove is $\angle XIM=\angle ZIN$ . From the condition $2BC=AB+AC$ we get that $XM=ZN$ . Thus $\triangle INZ= \triangle IMX$ . So we get that $\angle XIM=\angle ZIN$ , from which follows that $AMIN$ is cyclic.

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Babai

#7 h Jun 1, 2012, 5:41 AM • 2 Y

Y by Adventure10, Mango247

The problem is equivalent to prove that $IM=IN$ . i.e. $IM^2=IN^2$
Use barycentric coordinate and if we compare it remains to prove that $S_B(s-2b)=S_C(s-2c)$ where $S_B=\frac{1}{2}(a^2+c^2-b^2)$ and like wise $S_C$ .
If we calculate $\frac{S_B}{S_C}=\frac{5c-3b}{5b-3c}$ which is nothing but $\frac{s-2c}{s-2b}$ .
[While calculating we use the fact $a=\frac{b+c}{2}$ ]
Hence proved.

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#8 h Aug 27, 2017, 11:53 PM • 2 Y

Y by Projeto_Ramanujan, Adventure10

Open the gates

This post has been edited 1 time. Last edited by Gomes17, Aug 27, 2017, 11:54 PM
Reason: typo

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#9 h Aug 28, 2017, 2:32 AM • 2 Y

Y by Adventure10, Mango247

Let line AI intersect (ABC) at A and P. It follows that BP = CP = IP = r by the incenter excenter lemma.

Applying a homothety about A with scale factor 2 maps M to B and N to C. If AMIN were cyclic, the homothety would also map I to P. So it is sufficient to prove that AI = IP, or AP = 2r.

Let AB = c, AC = b, and BC = a. The problem's condition tells us that 2a = b+c. Applying Ptolemy's on ABPC gives:

$AB \cdot PC + BP \cdot AC = AP \cdot BC \rightarrow cr+br=a \cdot AN \rightarrow AN = 2r$ as desired.

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jayme

#10 h Aug 30, 2017, 1:38 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
I being the midpoint of AA'...we are done by homothety...

Sincerely
Jean-Louis

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Taco12

#11 h Jul 19, 2022, 7:07 PM • 2 Y

Y by RedFireTruck, megarnie

MAN IS MANI

We use barycentric coordinates, with $ABC$ as the reference triangle. We have $M = \left(\frac{1}{2}, \frac{1}{2}, 0\right), N = \left(\frac{1}{2}, 0, \frac{1}{2}\right), I = (a:b:c)$ . Consider the circumcircle of $AMN$ . We must show that $I$ also lies on this circle. Plugging in each of the points $A, M, N$ into the circle formula, we get $u=0, w=\frac{b^2}{2}, v=\frac{c^2}{2}$ . The equation of $(AMN)$ is $-a^2yz-b^2zx-c^2xy+\left(\frac{c^2y+b^2z}{2}\right)(x+y+z)=0.$ Now, plugging in $(a:b:c)$ , we have
$\begin{align*} -a^2bc-ab^2c-abc^2+\left(\frac{bc^2+b^2c}{2}\right)(x+y+z)=0 \\ abc(a+b+c)=\frac{(bc^2+b^2c)(a+b+c)}{2} \\ \frac{bc^2+b^2c}{2}=abc \\ a=\frac{1}{2}(b+c), \\ \end{align*}$ which is clear from the problem statement, so we are done. $\blacksquare$

This post has been edited 1 time. Last edited by Taco12, Jul 19, 2022, 8:17 PM

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#12 h Aug 19, 2022, 5:40 PM • 2 Y

Y by Mango247, Mango247

Literally same as @above, but posting for storage.

Clearly $M,N$ , the midpoints of $AB,AC,$ in addition to the circumcenter $O$ and $A$ are concyclic with diameter $OA.$ Thus, it suffices to show that $I$ lies on $(AMN)$ .

The general equation of a circle is $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$ for some constants $u,v,w$ . To find the constants $u,v,w$ which determine $\Omega$ , we can plug in each of $A,M,N$ in the general form and solve for $u,v,w$ . Note that the equation of a circle is homogenous so we can use homogenized coordinates here.

Plugging in $A=(1,0,0)$ gives $u=0$ .
Plugging in $M=(1:0:1)$ gives $u+w=\frac{b^2}2$ .
Plugging in $N=(1:1:0)$ gives $u+v=\frac{c^2}2$ .

Plugging in these constants in the general equation, we find that $\Omega$ has equation $-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{c^2}2\cdot y+\frac{b^2}2\cdot z\right)=0.$ Hence, using Barycentric Power of a Point, we can compute $\text{Pow}_{\Omega}(I)=-(a+b+c)(abc)+(a+b+c)\left(\frac{c^2}2\cdot b+\frac{b^2}2\cdot c\right)$ since $I=(a:b:c)$ . But this is $0$ since $a=\frac{b+c}2$ . Done.

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